In Fig. 16.201, ABCD is a quadrilateral inscribed in a circle with centre O. CD is produced to E such that ∠AED = 95° and ∠OBA = 30°. Find ∠OAC.
∠ADE = 95o (Given)
Since,
OA = OB, so
∠OAB = ∠OBA
∠OAB = 30o
∠ADE + ∠ADC = 180o (Linear pair)
95o + ∠ADC = 180o
∠ADC = 85o
We know that,
∠ADC = 2 ∠ADC
∠ADC = 2 * 85o
∠ADC = 170o
Since,
AO = OC (Radii of circle)
∠OAC = ∠OCA (Sides opposite to equal angle) (i)
In triangle OAC,
∠OAC + ∠OCA + ∠AOC = 180o
∠OAC + ∠OAC + 170o = 180o [From (i)]
2 ∠OAC = 10o
∠OAC = 5o
Thus,
∠OAC = 5o